An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

$A$ proton moving with a velocity of $2.5 \times 10^7 \, m/s$ enters a magnetic field of intensity $2.5 \, T$ making an angle of $30^o$ with the magnetic field. The force on the proton is:

An electron moves with speed $2 \times 10^5 \ m/s$ along the positive $x$-direction in the presence of a magnetic field of induction $B = \hat{i} + 4\hat{j} - 3\hat{k} \ T$. The magnitude of the force experienced by the electron in newtons is (Charge on the electron $= 1.6 \times 10^{-19} \ C$)

When a charged particle moving with velocity $\vec{V}$ is subjected to a magnetic field of induction $\vec{B}$,the force on it is non-zero. This implies that the:

An electron is orbiting in a circular orbit under the influence of a constant transverse magnetic field of strength $B$. Assuming that Bohr's postulate regarding the quantisation of angular momentum holds good for this electron,find the kinetic energy of the electron in the $n^{th}$ orbit. ($m_e =$ mass of electron,$e =$ magnitude of charge of electron)

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j}) \ \text{T}$ has an acceleration $(4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$. The value of $x$ is . . . . . . .